JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let the equation of the circle, which touches \(x\)-axis at the point \((a, 0), a\gt0\) and cuts off an intercept of length \(b\) on \(y\)-axis be \(x^2+y^2-\alpha x+\beta y+\gamma=0\). If the circle lies below \(x\)-axis, then the ordered pair \(\left(2 a, b^2\right)\) is equal to
- A \(\left(\gamma, \beta^2-4 \alpha\right)\)
- B \(\left(\alpha, \beta^2+4 \gamma\right)\)
- C \(\left(\gamma, \beta^2+4 \alpha\right)\)
- D \(\left(\alpha, \beta^2-4 \gamma\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\alpha, \beta^2-4 \gamma\right)\)
Step-by-step Solution
Detailed explanation
By pythagoras \(\mathrm{r}^2=\mathrm{a}^2+\frac{\mathrm{b}^2}{4}=\mathrm{P}^2\) \(r=\sqrt{\frac{4 a^2+b^2}{4}}\) Equation of circle is \((x-\alpha)^2+(y-\beta)^2=r^2\) \(x^2+y^2-2 a x-2 p y+\alpha^2+p^2-r^2=0\) comparision \(x^2+y^2-\alpha x+\beta y+r=0\)…
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